1. ## Bounded linear operator

Let X and Y be normed spaces and T_n : X--->Y (n=1,2.....) bounded linear operators. Show that converges T_n---->T implies that for every \epsilon > 0 there is an N such that for all n>N and all x in any given closed ball we have ||T_nx _ Tx||<\epsilon

2. Originally Posted by kinkong
Let X and Y be normed spaces and T_n : X--->Y (n=1,2.....) bounded linear operators. Show that converges T_n---->T implies that for every \epsilon > 0 there is an N such that for all n>N and all x in any given closed ball we have ||T_nx _ Tx||<\epsilon
What have you tried, man? So by definition what does it mean that $\displaystyle T_n\to T$ in the operator norm?

3. yes i tried....but i couldnt proceed

4. Originally Posted by kinkong
yes i tried....but i couldnt proceed
So, you're given some closed ball $\displaystyle B_\delta(x)$ and what you want to prove is that $\displaystyle \left\|T_ny-Ty\right\|\to 0$ for every [tex]y\in B_\delta(x)[tex], right? Well, you know that $\displaystyle \|y\|\leqslant \|x\|+\delta$ for every $\displaystyle y\in B_\delta(x)$ and so consequently $\displaystyle \left\|T_ny-T_y\right\|\leqslant \|T_n-T\|_\text{op}\|y\|\leqslant \|T_n-T\|(\|x\|+\delta)\to0$.